Answer:
A super hero because most people hate villains and I don't wanna be hated :)
Answer:
the rotational kinetic energy of the disk is 5,133.375 J
Explanation:
Given;
mass of the disk, m = 27 kg
radius of the disk, r = 1.3 m
angular speed, ω = 15 rad/s
The rotational kinetic energy of the disk is calculated as;

Therefore, the rotational kinetic energy of the disk is 5,133.375 J
Answer:
the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is <em>E)3.45 × 10⁸ m</em>
Explanation:
The force that the Earth exerts on a mass m is
F_e = (G M_e m) / R_e²
where
- G is the universal gravitational constant
- M_e is the mass of Earth
- R_e is the radius of Earth
The force that the Moon exerts on a mass m is
F_m = (G M_m m) / R_m²
where
- G is the universal gravitational constant
- M_m is the mass of the Moon
- R_m is the radius of the Moon
Therefore, the point where the gravitational pulls of Earth and Moon are equal is:
F_e = F_m
R_e + R_m = R = 3.84×10⁸ m
Thus,
(G M_e m) / R_e² = (G M_m m) / R_m²
M_e / R_e² = M_m / (R - R_e²)
(R - R_e²) / R_e² = M_m / M_e
(R - R_e) / R_e = (M_m / M_e)^1/2
R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2
R/ R_e = (M_m / M_e)^1/2 + 1
R_e = R / [(M_m / M_e)^1/2 + 1]
R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]
R_e = 3.45 × 10⁸ m
Therefore, the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is <em>3.45 × 10⁸ m.</em>
Divide 3500km by 3h (according to the equation v = d/t) and you get 1167km/h.
Answer:
5.3 seconds
Explanation:
Vx = Vx0 + Ax .T
A= de-acceleration
Vx =0 final at rest velocity
Vx0 = initial velocity
T= time to bring the object to a stop
therefore
0 = 52 m/s - (9.8 m/s^2) . T
-52 m/s/ -9.8 m/sec^2 =T
5.3 seconds= T